OpenStudy (anonymous):

Tommy flips a coin five times. How many ways can he get four heads and one tail?

3 years ago
OpenStudy (anonymous):

A 10 ways B 32 ways C 5 ways

3 years ago
OpenStudy (anonymous):

@ganeshie8

3 years ago
OpenStudy (acxbox22):

can you show me your attempt on solving this problem so I know where to start

3 years ago
OpenStudy (anonymous):

I don't know how to do this problem

3 years ago
OpenStudy (anonymous):

please help me

3 years ago
OpenStudy (anonymous):

like this...

3 years ago
OpenStudy (acxbox22):

so you can get HHHHT HHHTH HHTHH HTHHH THHHH I will give credit to hoblos as he aswered the same question like this

3 years ago
OpenStudy (anonymous):

|dw:1395596062854:dw| so 4 of the spots will be filled with heads and on with tails. how many of the spots could possibly be tails?

3 years ago
OpenStudy (anonymous):

1 spot

3 years ago
OpenStudy (anonymous):

well, for a particular sequence yes but any of the spots could be tails.

3 years ago
OpenStudy (anonymous):

ok

3 years ago
OpenStudy (anonymous):

is there a possible for it to be 10 ways

3 years ago
OpenStudy (anonymous):

the way to think of this is witth combinations. you have 5 availble spots, choose 1. that gives \[_5C_1 =\left(\begin{matrix}5 \\ 1\end{matrix}\right)=\frac{ 5! }{ (5-1)!\,\,1! }=5\]

3 years ago
OpenStudy (anonymous):

ok thank you

3 years ago
OpenStudy (anonymous):

well, for 2 tails out of 5 flips you'd get \[_5C_2 = \left(\begin{matrix}5 \\ 2\end{matrix}\right)=\frac{ 5! }{ (5-2)!\,\,2! }=10\]

3 years ago